Chemistry: Atoms First 2e · Transition Metals and Coordination Chemistry
Spectroscopic and Magnetic Properties of Coordination Compounds
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In 30 seconds
Why is a solution of copper(II) sulfate blue, a ruby red, and an emerald green? The answer lies in the d orbitals of the transition-metal ion and how they respond to the electric field of the surrounding ligands. When ligands approach a metal ion, they break the symmetry of the free ion and split the five d orbitals into groups of different energy — the crystal field splitting. Electrons can then absorb visible light and jump from a lower-energy d orbital to a higher one, and the wavelength absorbed (which the eye sees as the complementary color) reports the size of the splitting. This is the crystal field theory, and it explains not only color but also magnetism: complexes with unpaired d electrons are Paramagnetic Attracted into a magnetic field because of unpaired electrons (attracted into a magnetic field), while complexes with all electrons paired are Diamagnetic Weakly repelled by a magnetic field because all electrons are paired Full entry → (weakly repelled).
This topic connects the coordination chemistry of the previous topic to real observable properties. Given a metal, its oxidation state, and its ligands, you can predict the number of unpaired electrons, whether the complex is high spin or low spin, roughly what color it will be, and how strongly it responds to a magnet — skills that matter in gemology, analytical chemistry, biochemistry (hemoglobin's color change on oxygen binding!), and MRI contrast design.
Why this matters
- Color in the real world: gems (ruby = Cr3+ in alumina; emerald = Cr3+ in beryl), pigments, and dyes owe their colors to d-d transitions in transition-metal complexes.
- Medicine and biology: hemoglobin turns red when oxygenated and darker red-purple when deoxygenated — a spectroscopic change used in pulse oximetry; MRI contrast agents are paramagnetic gadolinium complexes.
- Analytical chemistry: UV-visible spectrophotometry quantifies metal ions by measuring the light absorbed by their complexes; the color is a built-in concentration probe.
- Materials: the magnetic properties of transition-metal complexes underlie data storage, MRI, and magnetic separation technologies.
- Exams: predicting high spin vs. low spin, counting unpaired electrons, and relating Δ (crystal field splitting) to color are classic problems.
The college version
Core Concepts
Crystal field theory: why d orbitals split
In a free (gas-phase) transition-metal ion, all five d orbitals have the same energy. When six ligands approach octahedrally, they point along the x, y, and z axes, directly at the dx2-y2 and dz2 orbitals (the eg set), raising those orbitals in energy. The dxy, dxz, and dyz orbitals (the t2g set) point between the axes, away from the ligands, so they stay lower in energy. The energy gap between the two sets is the crystal field splitting energy, Δoct:
\[\Delta{\text{oct}} = E{eg} - E{t_{2g}}\]
For an octahedral complex, Δoct is typically in the range of 100–400 kJ/mol — exactly the energy of visible-light photons, which is why these complexes are colored.
Factors that control Δ
Three factors set the size of the splitting:
- The ligand: ligands arranged by their ability to split d orbitals form the Spectrochemical series The ranking of ligands by their ability to split d orbitals Full entry → — roughly, I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < NO2- < CN- < CO. Strong-field ligands (right end) give large Δ; weak-field ligands (left end) give small Δ.
- The metal's oxidation state: higher oxidation states pull ligands closer, increasing Δ (e.g., [Fe(CN)6]3- with Fe3+ has a larger Δ than [Fe(CN)6]4- with Fe2+).
- The metal's position in the periodic table: Δ increases going down a group (4d > 3d) and across a period.
High spin vs. low spin
When electrons fill the split d orbitals, two competing rules apply: Hund's rule says spread electrons out (unpaired) for stability, but a large Δ makes pairing electrons in the lower set energetically favorable. The outcome for d4–d7 ions depends on Δ:
- Weak field (small Δ): electrons spread out first — high spin (maximum unpaired electrons). Example: [Fe(H2O)6]2+ (d6) has four unpaired electrons.
- Strong field (large Δ): electrons pair up in the lower t2g set — low spin (minimum unpaired electrons). Example: [Fe(CN)6]4- (d6) has zero unpaired electrons.
The d1, d2, d3, d8, d9, and d10 cases have only one possible arrangement (no choice), so only d4–d7 complexes show high/low spin behavior.
Color: absorption and complementary colors
A complex appears colored because it absorbs some wavelengths of visible light and transmits/reflects the rest; the eye sees the complementary color of the absorbed light. Larger Δ means higher-energy (shorter-wavelength) light is absorbed:
- [Ti(H2O)6]3+ (d1) absorbs green-yellow light (~500 nm) and looks violet.
- [Cu(H2O)4]2+ absorbs red-orange light and looks blue.
- [Co(NH3)6]3+ (strong field, large Δ) absorbs higher-energy light than [CoF6]3- (weak field, small Δ), so the ammine complex looks different in color from the fluoride complex.
The relationship between the absorbed photon's energy and wavelength is:
\[E = \frac{hc}{\lambda}\]
Magnetism: unpaired electrons respond to a field
- Paramagnetic complexes have one or more unpaired d electrons; the electron spins act like tiny magnets, so the complex is attracted into a magnetic field.
- Diamagnetic complexes have all electrons paired and are weakly repelled.
The magnetic moment measures the strength of the response. For a complex with n unpaired electrons:
\[\mu = \sqrt{n(n+2)}\ \text{BM}\]
where BM is the Bohr magneton. Measuring μ experimentally tells you n — a classic way to distinguish high-spin from low-spin complexes.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Absorbed color | Observed color | A complex absorbs one wavelength and shows its complement; violet-looking [Ti(H2O)6]3+ actually absorbs green-yellow |
| High spin | Low spin | High spin = weak field, small Δ, more unpaired electrons; low spin = strong field, large Δ, fewer unpaired electrons |
| d4–d7 spin choice | All d counts having a spin choice | Only d4–d7 can be high or low spin; d1–d3, d8–d10 have a single possible arrangement |
| Paramagnetic | Ferromagnetic | Paramagnetic complexes respond only in an external field; ferromagnetic materials (iron metal) retain magnetization — different phenomena |
| Δoct | Δtet | Tetrahedral splitting is much smaller (~4/9 of octahedral), so tetrahedral complexes are almost always high spin |
| Crystal field splitting being caused by bonding | Splitting being caused by ligand charge alone | Crystal field theory treats ligands as point charges/dipoles (electrostatic); the reality includes covalence — but the theory successfully explains spectra and magnetism |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Picture five equal parking spots for electrons around a metal atom. When ligands arrive, they turn two of the spots into premium, high-up spots and leave three spots low. Electrons like the low spots. If the gap between low and high spots is small, electrons spread out and stay single (high spin, attracted to magnets); if the gap is huge, they'd rather share the low spots in pairs (low spin, not magnetic). When an electron jumps from a low spot to a high spot, it swallows a specific color of light — and whatever color is left over is the color you see.
Worked example
Example 1: Energy of an absorbed photon — why [Ti(H2O)6]³⁺ is violet
The ion [Ti(H2O)6]3+ (d1) absorbs light at 500 nm. Calculate the energy of one absorbed photon, then the energy per mole, and identify what color the solution appears.
Step 1 — Write the formula before substituting:
\[E = \frac{hc}{\lambda}\]
Step 2 — Substitute values. h = 6.626 × 10-34 J s, c = 3.00 × 108 m/s, λ= 500 nm = 500 × 10-9 m:
\[E = \frac{(6.626 \times 10^{-34}\ \text{J s})(3.00 \times 10^8\ \text{m s}^{-1})}{500 \times 10^{-9}\ \text{m}} = 3.98 \times 10^{-19}\ \text{J}\]
Step 3 — Per mole, using Avogadro's number:
\[E_{\text{per mol}} = (3.98 \times 10^{-19}\ \text{J})(6.022 \times 10^{23}\ \text{mol}^{-1}) = 2.40 \times 10^5\ \text{J/mol} = 240\ \text{kJ/mol}\]
Step 4 — Color. 500 nm (green-yellow) is absorbed; the complementary color is violet, so the solution looks violet. Note that 240 kJ/mol sits inside the 100–400 kJ/mol range quoted for Δoct — the photon energy and the d-orbital gap match.
Example 2: High spin or low spin — and magnetism — for two iron(II) complexes
[Fe(H2O)6]2+ and [Fe(CN)6]4- are both d6 octahedral complexes. Predict the electron arrangement and magnetic behavior of each.
Step 1 — Place the ligands in the spectrochemical series. H2O is a weak-field ligand; CN- is a strong-field ligand.
Step 2 — High spin for the weak field. For [Fe(H2O)6]2+, Δ is small, so electrons spread out per Hund's rule: t2g4 eg2 — four unpaired electrons, paramagnetic.
Step 3 — Low spin for the strong field. For [Fe(CN)6]4-, Δ is large, so electrons pair in the lower set: t2g6 eg0 — zero unpaired electrons, diamagnetic.
Step 4 — Magnetic moments for comparison (using μ= n(n+2) BM):
\[\mu_{[Fe(H_2O)6]^{2+}} = \sqrt{4(4+2)} = \sqrt{24} = 4.9\ \text{BM} \qquad \mu{[Fe(CN)_6]^{4-}} = \sqrt{0(0+2)} = 0\ \text{BM}\]
A measured moment near 4.9 BM confirms the high-spin assignment; a moment of ~0 confirms low spin. This is exactly how experimentalists tell the two apart.
Example 3: Predicting the number of unpaired electrons for a d⁵ ion
[Mn(CN)6]3- contains Mn3+ (d4). Is it high spin or low spin, and what magnetic moment do you predict?
Step 1 — Ligand field strength. CN- is a strong-field ligand → large Δ → low spin.
Step 2 — Fill the orbitals. d4 low spin: all four electrons enter the lower t2g set, pairing two of them: t2g4 eg0 — two unpaired electrons.
Step 3 — Magnetic moment:
\[\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.8\ \text{BM}\]
The same d4 ion with a weak-field ligand (e.g., [Mn(H2O)6]3+) would be high spin: t2g3 eg1, four unpaired electrons, μ= 4.9 BM. The ligand alone decides which spin state you observe.
Key takeaways
- Octahedral splitting: 5 d orbitals → t2g (lower, 3 orbitals) + eg (higher, 2 orbitals); gap = Δoct.
- Δoct values (100–400 kJ/mol) match visible-light photon energies → colored complexes.
- Spectrochemical series (partial): I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < NO2- < CN- < CO.
- High spin = weak field = small Δ = maximum unpaired electrons; low spin = strong field = large Δ = electrons pair up (only d4–d7).
- A complex absorbs one color and shows its complementary color; larger Δ → higher-energy absorbed light → different observed color.
- E = hc/λ connects absorbed wavelength to photon energy; μ= n(n+2) BM connects unpaired electrons to magnetic moment.
- Paramagnetic = unpaired electrons (attracted); diamagnetic = all paired (weakly repelled).
- [Fe(CN)6]4- (CN⁻, strong field, d6): low spin, 0 unpaired, diamagnetic. [Fe(H2O)6]2+ (H2O, weak field, d6): high spin, 4 unpaired, paramagnetic.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Why does an octahedral ligand field split the five d orbitals into two sets, and what are the sets called?
Show answer
Six ligands approaching along the x, y, z axes point directly at dx2-y2 and dz2, raising them (the eg set); the dxy, dxz, dyz orbitals point between the axes and stay lower (the t2g set).
Arrange these ligands in order of increasing field strength: CN-, Cl-, NH3, H2O.
Show answer
Cl- < H2O < NH3 < CN- (weak to strong field).
Is [Co(NH3)6]3+ (d6) high spin or low spin? How many unpaired electrons does it have?
Show answer
NH3 is a strong-field ligand → low spin: t2g6 eg0, zero unpaired electrons, diamagnetic.
A complex absorbs light at 600 nm. Calculate the photon energy in J, and state roughly what color the solution appears (given that 600 nm is orange-red).
Show answer
E = hc/λ= (6.626 × 10-34)(3.00 × 108)/(600 × 10-9) = 3.31 × 10-19 J. Absorbing orange-red, the solution appears its complement — blue-green.
Explain why [Fe(H2O)6]2+ is paramagnetic but [Fe(CN)6]4- is diamagnetic, even though both are d6.
Show answer
H2O is a weak-field ligand → small Δ → high spin t2g4 eg2 (4 unpaired, paramagnetic). CN- is strong field → large Δ → low spin t2g6 eg0 (0 unpaired, diamagnetic).
What magnetic moment do you predict for a d5 high-spin complex?
Show answer
High-spin d5: five unpaired electrons → μ= 5(5+2) = 35 = 5.9 BM.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Crystal field splitting (Δ)
- The energy gap between d-orbital groups created by ligand electric fields
- t2g / eg orbitals
- The lower-energy (3) / higher-energy (2) d-orbital sets in octahedral complexes
- Spectrochemical series
- The ranking of ligands by their ability to split d orbitals
- High spin / low spin
- Electron arrangements with maximum / minimum unpaired electrons (d4–d7)
- Paramagnetic
- Attracted into a magnetic field because of unpaired electrons
- Diamagnetic
- Weakly repelled by a magnetic field because all electrons are paired
- Magnetic moment (μ)
- A measured quantity, n(n+2) BM, reporting the number of unpaired electrons
- Complementary color
- The color seen when a complex absorbs its opposite color on the color wheel
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
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